How do you determine the equation of a line perpendicular to y = 4x + 1?

To determine the equation of a line perpendicular to \( y = 4x + 1 \), use the negative reciprocal of the original slope.

In more detail, the slope (or gradient) of the given line \( y = 4x + 1 \) is 4. For a line to be perpendicular to this, its slope must be the negative reciprocal of 4. The negative reciprocal of 4 is \(-\frac{1}{4}\). This means the slope of the perpendicular line will be \(-\frac{1}{4}\).

Next, we need to use the point-slope form of a line equation, which is \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \((x_1, y_1)\) is a point on the line. If you have a specific point through which the perpendicular line passes, substitute that point and the slope \(-\frac{1}{4}\) into the equation.

For example, if the perpendicular line passes through the point (2, 3), substitute \( m = -\frac{1}{4} \), \( x_1 = 2 \), and \( y_1 = 3 \) into the point-slope form:
\[ y - 3 = -\frac{1}{4}(x - 2) \]

Simplify this to get the equation in the slope-intercept form \( y = mx + c \):
\[ y - 3 = -\frac{1}{4}x + \frac{1}{2} \]
\[ y = -\frac{1}{4}x + \frac{1}{2} + 3 \]
\[ y = -\frac{1}{4}x + \frac{7}{2} \]

So, the equation of the line perpendicular to \( y = 4x + 1 \) and passing through (2, 3) is \( y = -\frac{1}{4}x + \frac{7}{2} \).